x varies from a to a + h, the function f varies from f (a) to f (a + h), then we
know that the difference quotient
is the average rate of change of f over the interval from a to a + h.
Thus, the average velocity of a moving object over some time interval is the
change in distance divided by the change in time, the average rate of growth of a
colony of fruit flies over some interval of time is the change in size of the colony
divided by the time elapsed, the average rate of change in the profit of a
company on some gadget with respect to production is the change in profit
divided by the change in the number of gadgets produced.
The (instantaneous) rate of change of f at a, or the derivative of f at a, is the
limit of the average rate of change as h → 0:
On the graph of y = f(x), the rate at which the y-coordinate changes with
respect to the x-coordinate is f ′(x), the slope of the curve. The rate at which s(t),
the distance traveled by a particle in t seconds, changes with respect to time is s
′(t), the velocity of the particle; the rate at which a manufacturer’s profit P(x)
changes relative to the production level x is P′(x).
Example 4 __
Let G = 400(15 − t)^2 be the number of gallons of water in a cistern t minutes
after an outlet pipe is opened. Find the average rate of drainage during the first 5
minutes and the rate at which the water is running out at the end of 5 minutes.
SOLUTION: The average rate of change during the first 5 min equals
The average rate of drainage during the first 5 min is 10,000 gal/min.
The instantaneous rate of change at t = 5 is G′(5). Since
G′(t) = −800(15 − t),G′(5) = −800(10) = −8000 gal/min. Thus the rate of drainage at the end of 5 min
is 8000 gal/min.